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Found in: Page 442

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Solve the differential equation ${\mathbf{f}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(t\right){\mathbf{+}}{\mathbf{f}}{\mathbf{\text{'}}}\left(t\right){\mathbf{-}}{\mathbf{12}}{\mathbf{f}}\left(t\right){\mathbf{=}}{\mathbf{0}}$and find solution of the differential equation.

The solution is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-4\mathrm{t}}$.

See the step by step solution

## Step1: Definition of characteristic polynomial

Consider the linear differential operator

$\mathrm{T}\left(\mathrm{f}\right)={\mathrm{f}}^{\left(\mathrm{n}\right)}+{\mathrm{a}}_{\mathrm{n}-1}{\mathrm{f}}^{\left(\mathrm{n}-1\right)}+...+{\mathrm{a}}_{1}\mathrm{f}\text{'}+{\mathrm{a}}_{0}\mathrm{f}$.

The characteristic polynomial of T is defined as

${\mathrm{p}}_{\mathrm{t}}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^{\mathrm{n}}+{\mathrm{a}}_{\mathrm{n}-1}\mathrm{\lambda }+...{\mathrm{a}}_{1}\mathrm{\lambda }+{\mathrm{a}}_{0}$.

## Step2: Determination of the solution

The characteristic polynomial of the operator as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}\text{'}+\mathrm{f}\text{'}-12$

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^{2}+\mathrm{\lambda }‐12$

Solve the characteristic polynomial and find the roots as follows.

$\lambda =\frac{‐b±\sqrt{{b}^{2}‐4ac}}{2a}\phantom{\rule{0ex}{0ex}}\lambda =\frac{‐1±\sqrt{{\left(1\right)}^{2}‐4\left(1\right)\left(‐12\right)}}{2\left(1\right)}\left\{\because a=1,,b=1,c=‐12\right\}\phantom{\rule{0ex}{0ex}}\lambda =\frac{‐1±\sqrt{1+48}}{2}\phantom{\rule{0ex}{0ex}}\lambda =\frac{‐1±\sqrt{49}}{2}$

Simplify further as follows.

$\lambda =\frac{‐1±7}{2}\phantom{\rule{0ex}{0ex}}\lambda =\frac{‐1±7}{2},\lambda =\frac{‐1‐7}{2}\phantom{\rule{0ex}{0ex}}\lambda =\frac{6}{2},\lambda =\frac{‐8}{2}\phantom{\rule{0ex}{0ex}}\lambda =3,‐4$

Therefore, the roots of the characteristic polynomials are 3 and $‐4$ .

## Step3: Conclusion of the solution

Since, the roots of the characteristic equation is different real numbers.

The exponential functions ${\mathrm{e}}^{3\mathrm{t}}$ and ${\mathrm{e}}^{-4\mathrm{t}}$ form a basis of the kernel of T .

Hence, they form a basis of the solution space of the homogenous differential equation is $\mathrm{T}\left(\mathrm{f}\right)=0$.

Thus, the general solution of the differential equation $\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)+\mathrm{f}\text{'}\left(\mathrm{t}\right)‐12=0$ is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-4\mathrm{t}}$.